The upper chamber of an hour-glass is a cone of radius 3 inches and height 10 inches and, if full, it requires exactly one hour to empty. Assuming that the sand falls through the aperture at a constant rate, how fast is the level falling when: a) the depth of the sand is 6 inches? b) 52.5 minutes have elapsed from the time when the hour-glass was full in the upper chamber?

Question

The upper chamber of an hour-glass is a cone of radius 3 inches and height 10 inches and, if full, it requires exactly one hour to empty.  Assuming that the sand falls through the aperture at a constant rate, how fast is the level falling when: a)  the depth of the sand is 6 inches? b)  52.5 minutes have elapsed from the time when the hour-glass was full in       the upper chamber?

Help me! I don't know What should I do. I tried one's best but my answer always are wrong

Steve · Answer

So, what did you try? It'd be nice if you showed your work on the problem. http://www.jiskha.com/display.cgi?id=1476782673

ZEN · Answer

The answer are 9.26 in/hr and 13.33 in/hr

Steve · Answer

Well, I made a boo-boo. Clearly you did not bother to read my work, or you would have found it.

I was just as blind, because if the glass empties 10 in. in one hour, clearly the .15 in/hr I got for an answer was way too small.

So, it all goes back to this: I swapped y and r in the formula. So, doing it right,

v = π/3 r^2 y = 3π/100 y^3

dv/dt = 9π/100 y^2 dy/dt
-π/2 = 9π/100 * 6^2 dy/dt
dy/dt = -25/162 in/min
Or, -9.26 in/hr
Since it is negative, the level is falling at 9.26 in/hr